MHT CET 2019 Mathematics Question Paper with Answer and Solution

(A) The given pair of lines are $xy = 0$ and $xy + 5x - 4y - 20 = 0$.
From $xy = 0$,we get the lines $x = 0$ (y-axis) and $y = 0$ (x-axis).
From the second equation,$xy + 5x - 4y - 20 = 0$,we can factorize it as:
$x(y + 5) - 4(y + 5) = 0$
$(x - 4)(y + 5) = 0$
This gives the lines $x = 4$ and $y = -5$.
The region is bounded by the lines $x = 0$,$x = 4$,$y = 0$,and $y = -5$.
This forms a rectangle with length $|4 - 0| = 4$ units and width $|0 - (-5)| = 5$ units.
The area of the rectangle is $\text{length} \times \text{width} = 4 \times 5 = 20$ square units.

(B) We have the equation of the line $y = x$ and the equation of the circle $x^2 + y^2 - 2x = 0$.
To find the intersection points of the given line and circle,substitute $y = x$ into the circle equation:
$x^2 + x^2 - 2x = 0$
$2x^2 - 2x = 0$
$2x(x - 1) = 0$
$x = 0, 1$
When $x = 0$,$y = 0$; when $x = 1$,$y = 1$.
Thus,the coordinates of the endpoints of the diameter $AB$ are $(0, 0)$ and $(1, 1)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the points $(0, 0)$ and $(1, 1)$:
$(x - 0)(x - 1) + (y - 0)(y - 1) = 0$
$x(x - 1) + y(y - 1) = 0$
$x^2 - x + y^2 - y = 0$
$x^2 + y^2 - x - y = 0$

(A) The given equation of the circle is $x^2+y^2-6x-4y-12=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2$.
The center of the circle is $(-g, -f) = (3, 2)$.
$A$ circle concentric with the given circle will have the same center $(3, 2)$.
Let the equation of the required circle be $(x-3)^2+(y-2)^2=r^2$.
Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coordinate of the center,so $r = |3| = 3$.
Substituting $r=3$ into the equation: $(x-3)^2+(y-2)^2=3^2$.
Expanding this,we get $x^2-6x+9+y^2-4y+4=9$.
Simplifying,we get $x^2+y^2-6x-4y+4=0$.

(B) Given,the length of the latus rectum is $\frac{2b^2}{a} = \frac{18}{5}$ $\Rightarrow \frac{b^2}{a} = \frac{9}{5}$ $\Rightarrow b^2 = \frac{9}{5}a \dots (i)$.
Given,eccentricity $e = \frac{4}{5}$.
We know that $e^2 = 1 - \frac{b^2}{a^2}$,so $\frac{16}{25} = 1 - \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = 1 - \frac{16}{25} = \frac{9}{25}$.
Substituting $b^2 = \frac{9}{5}a$ into the equation: $\frac{\frac{9}{5}a}{a^2} = \frac{9}{25}$ $\Rightarrow \frac{9}{5a} = \frac{9}{25}$ $\Rightarrow a = 5$.
Now,$b^2 = \frac{9}{5}(5) = 9$.
Thus,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

(A) The given equation of the hyperbola is $25x^2 - 9y^2 = 225$.
Dividing both sides by $225$,we get:
$\frac{25x^2}{225} - \frac{9y^2}{225} = 1$
$\frac{x^2}{9} - \frac{y^2}{25} = 1$.
Comparing this with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we find $a^2 = 9$ and $b^2 = 25$.
The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting the values,we get:
$e = \sqrt{1 + \frac{25}{9}} = \sqrt{\frac{9 + 25}{9}} = \sqrt{\frac{34}{9}} = \frac{\sqrt{34}}{3}$.

(B) Given that the length of the transverse axis is $2a = 6$,so $a = 3$.
Given that the length of the latus rectum is $\frac{2b^2}{a} = \frac{8}{3}$.
Substituting $a = 3$ into the equation: $\frac{2b^2}{3} = \frac{8}{3} \implies 2b^2 = 8 \implies b^2 = 4$.
The standard equation of a hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 9$ and $b^2 = 4$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Multiplying by $36$,we get $4x^2 - 9y^2 = 36$.

(B) The given hyperbola is $x^2 - y^2 = a^2$,which is a rectangular hyperbola with eccentricity $e = \sqrt{2}$.
The foci are $S(ae, 0) = (a\sqrt{2}, 0)$ and $S'(-ae, 0) = (-a\sqrt{2}, 0)$.
For any point $P(x_1, y_1)$ on the hyperbola,the focal distances are given by $SP = |ex_1 - a|$ and $S'P = |ex_1 + a|$.
Since $e = \sqrt{2}$,we have $SP = |\sqrt{2}x_1 - a|$ and $S'P = |\sqrt{2}x_1 + a|$.
Therefore,$SP \cdot S'P = |(\sqrt{2}x_1 - a)(\sqrt{2}x_1 + a)| = |2x_1^2 - a^2|$.
Since $P(x_1, y_1)$ lies on $x^2 - y^2 = a^2$,we have $x_1^2 - y_1^2 = a^2$,which implies $x_1^2 = a^2 + y_1^2$.
Substituting this into the expression: $SP \cdot S'P = |2(a^2 + y_1^2) - a^2| = |a^2 + 2y_1^2|$.
However,for a rectangular hyperbola,the product of focal distances $SP \cdot S'P$ is equal to the square of the distance from the center to the point $P$,which is $x_1^2 + y_1^2$.

(B) tautology is a statement pattern that is always true for all possible truth values of its components.
Check option $(b): p \rightarrow (q \vee p)$
$= \sim p \vee (q \vee p)$
$= (\sim p \vee p) \vee q$
$= T \vee q$
$= T$
Since the result is always true $(T)$,the statement pattern $p \rightarrow (q \vee p)$ is a tautology.

(D) Let $p$ be the statement "Two triangles are congruent" and $q$ be the statement "Their areas are equal". The given statement is $p \rightarrow q$.
The inverse of $p \rightarrow q$ is $\sim p \rightarrow \sim q$,which is: "If two triangles are not congruent,then their areas are not equal."
The contrapositive of the inverse $(\sim p \rightarrow \sim q)$ is $\sim(\sim q) \rightarrow \sim(\sim p)$,which simplifies to $q \rightarrow p$. This is: "If the areas of two triangles are equal,then they are congruent."

(A) Given: $p = T, q = T, r = F, s = F$.
For $a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$
Substitute the values: $\sim(T \wedge \sim F) \vee(\sim T \vee F)$
$= \sim(T \wedge T) \vee(F \vee F)$
$= \sim(T) \vee(F)$
$= F \vee F = F$.
For $b: (p \vee s) \leftrightarrow(q \wedge r)$
Substitute the values: $(T \vee F) \leftrightarrow(T \wedge F)$
$= (T) \leftrightarrow(F)$
Since the truth values are different,the biconditional statement is $F$.
Therefore,the truth values of $a$ and $b$ are $F, F$.

(C) Key Idea: $A$ statement that is neither a tautology nor a contradiction is a contingency.
Option $A$: $(p \vee q) \vee \sim q \equiv p \vee (q \vee \sim q) \equiv p \vee T \equiv T$. This is a tautology.
Option $B$: $(p \vee q) \vee \sim p \equiv (p \vee \sim p) \vee q \equiv T \vee q \equiv T$. This is a tautology.
Option $C$: $(p \vee q) \wedge \sim q \equiv (p \wedge \sim q) \vee (q \wedge \sim q) \equiv (p \wedge \sim q) \vee F \equiv p \wedge \sim q$.
If $p=T, q=T$,then $p \wedge \sim q = F$.
If $p=T, q=F$,then $p \wedge \sim q = T$.
Since the truth value depends on $p$ and $q$,it is a contingency.
Option $D$: $p \rightarrow (p \vee q) \equiv \sim p \vee (p \vee q) \equiv (\sim p \vee p) \vee q \equiv T \vee q \equiv T$. This is a tautology.
Therefore,the correct statement is $(p \vee q) \wedge \sim q$.

(A) Key Idea: The negation of the universal quantifier $\forall$ (for all) is the existential quantifier $\exists$ (there exists),and the negation of the inequality $>$ is $\leq$.
Given the statement: $\forall n \in N, n+7 > 6$.
Applying the rules of negation:
$1$. Replace $\forall$ with $\exists$.
$2$. Negate the condition $n+7 > 6$,which becomes $n+7 \leq 6$.
Therefore,the negation of the given statement is $\exists n \in N$,such that $n+7 \leq 6$.

(C) Given: $p = T, q = T, r = F, s = F$.
Check Option $(A): (q \wedge r) \vee (\sim p \wedge s) \equiv (T \wedge F) \vee (F \wedge F) \equiv F \vee F \equiv F$.
Check Option $(B): (\sim p$ $\rightarrow q)$ $\rightarrow (r \wedge s) \equiv (\sim T$ $\rightarrow T)$ $\rightarrow (F \wedge F) \equiv (F$ $\rightarrow T)$ $\rightarrow F \equiv T$ $\rightarrow F \equiv F$.
Check Option $(C): (p$ $\rightarrow q) \vee (r \leftrightarrow s) \equiv (T$ $\rightarrow T) \vee (F \leftrightarrow F) \equiv T \vee T \equiv T$.
Check Option $(D): (p \wedge \sim r) \wedge (\sim q \vee s) \equiv (T \wedge \sim F) \wedge (\sim T \vee F) \equiv (T \wedge T) \wedge (F \vee F) \equiv T \wedge F \equiv F$.
Thus,option $(C)$ is true.

(A) We know that the negation of an implication is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Applying this rule to the given expression:
$\sim(p \rightarrow \sim q) \equiv p \wedge \sim(\sim q)$.
Using the law of double negation,$\sim(\sim q) \equiv q$.
Therefore,the expression simplifies to $p \wedge q$.

(C) The logical implication $p \rightarrow q$ can be expressed in several equivalent ways:
$1.$ If $p$ then $q$.
$2.$ $p$ only if $q$.
$3.$ $q$ is necessary for $p$.
$4.$ $p$ is sufficient for $q$.
Comparing these with the given options,option $C$ ($q$ only if $p$) is equivalent to $q \rightarrow p$,which is the converse of $p \rightarrow q$. Therefore,it is not equivalent to $p \rightarrow q$.

(B) Let the given statement be $S = (p \wedge q) \wedge [\sim r \vee (p \wedge q)] \vee (\sim p \wedge q)$.
Using the absorption law,$(p \wedge q) \wedge [\sim r \vee (p \wedge q)]$ simplifies to $(p \wedge q)$.
Thus,the expression becomes $S = (p \wedge q) \vee (\sim p \wedge q)$.
Using the distributive law,we can factor out $q$:
$S = (p \vee \sim p) \wedge q$.
Since $(p \vee \sim p)$ is a tautology $(T)$,
$S = T \wedge q = q$.
Therefore,the statement pattern is equivalent to $q$.

(A) The slopes of the two lines are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{1}{1+\sqrt{2}}$.
Rationalizing $m_2$: $m_2 = \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \sqrt{2}-1$.
The joint equation of lines passing through the origin with slopes $m_1$ and $m_2$ is given by $(y-m_1 x)(y-m_2 x) = 0$,which simplifies to $y^2 - (m_1+m_2)xy + m_1 m_2 x^2 = 0$.
Calculate the sum of slopes: $m_1+m_2 = (1+\sqrt{2}) + (\sqrt{2}-1) = 2\sqrt{2}$.
Calculate the product of slopes: $m_1 m_2 = (1+\sqrt{2})(\sqrt{2}-1) = (\sqrt{2})^2 - 1^2 = 2-1 = 1$.
Substituting these values into the equation: $y^2 - (2\sqrt{2})xy + 1x^2 = 0$,which is $x^2 - 2\sqrt{2}xy + y^2 = 0$.

(A) The slopes of the lines passing through the origin are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{-1}{1+\sqrt{2}}$.
Since $m_2 = \frac{-1}{1+\sqrt{2}} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{-(\sqrt{2}-1)}{2-1} = 1-\sqrt{2}$,the lines are $y = (1+\sqrt{2})x$ and $y = (1-\sqrt{2})x$.
Rearranging,we get $(y - (1+\sqrt{2})x) = 0$ and $(y - (1-\sqrt{2})x) = 0$.
The joint equation is $(y - (1+\sqrt{2})x)(y - (1-\sqrt{2})x) = 0$.
Expanding this,we get $y^2 - (1-\sqrt{2})xy - (1+\sqrt{2})xy + (1+\sqrt{2})(1-\sqrt{2})x^2 = 0$.
$y^2 - (1-\sqrt{2}+1+\sqrt{2})xy + (1-2)x^2 = 0$.
$y^2 - 2xy - x^2 = 0$.
Multiplying by $-1$,we get $x^2 + 2xy - y^2 = 0$.

(C) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
If these lines are perpendicular to each other,the condition is $a + b = 0$.
Given the equation $(1+\sin^2 \theta) x^2 + 2hxy + 2\sin \theta y^2 = 0$,we identify $a = 1+\sin^2 \theta$ and $b = 2\sin \theta$.
Applying the condition $a + b = 0$:
$(1+\sin^2 \theta) + 2\sin \theta = 0$
$(1+\sin \theta)^2 = 0$
$1+\sin \theta = 0$
$\sin \theta = -1$
For $\theta \in [0, 2\pi]$,the value of $\theta$ satisfying $\sin \theta = -1$ is $\theta = \frac{3\pi}{2}$.

(A) The given equation of the pair of lines is $x^2-4pxy+8y^2=0$.
This is in the form $ax^2+2hxy+by^2=0$,where $a=1$,$2h=-4p$,and $b=8$.
Let $m_1$ and $m_2$ be the slopes of the lines.
We know that $m_1+m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.
Substituting the values,$m_1+m_2 = -\frac{-4p}{8} = \frac{4p}{8} = \frac{p}{2}$ and $m_1m_2 = \frac{1}{8}$.
According to the problem,the sum of the slopes is three times their product:
$m_1+m_2 = 3(m_1m_2)$
$\frac{p}{2} = 3 \times \frac{1}{8}$
$\frac{p}{2} = \frac{3}{8}$
$p = \frac{3}{8} \times 2 = \frac{3}{4}$.

(D) The total number of ways to draw $3$ cards from a pack of $52$ cards is given by ${}^{52}C_3$.
There are $26$ red cards in a pack of $52$. The number of ways to draw $3$ red cards is ${}^{26}C_3$.
$\text{Required probability} = \frac{{}^{26}C_3}{{}^{52}C_3} = \frac{\frac{26 \times 25 \times 24}{3 \times 2 \times 1}}{\frac{52 \times 51 \times 50}{3 \times 2 \times 1}} = \frac{26 \times 25 \times 24}{52 \times 51 \times 50}$.
Simplifying the expression: $\frac{26}{52} \times \frac{25}{50} \times \frac{24}{51} = \frac{1}{2} \times \frac{1}{2} \times \frac{8}{17} = \frac{8}{68} = \frac{2}{17}$.

(B) When three dice are thrown,the total number of outcomes is $n(S) = 6^3 = 216$.
Let $E$ be the event that the sum of the numbers on the uppermost faces is less than $5$.
The possible outcomes for the sum being less than $5$ are:
$(1, 1, 1)$ (sum $= 3$)
$(1, 1, 2), (1, 2, 1), (2, 1, 1)$ (sum $= 4$)
Therefore,the number of favorable outcomes for the sum being less than $5$ is $n(E) = 1 + 3 = 4$.
The probability of the sum being less than $5$ is $P(E) = \frac{n(E)}{n(S)} = \frac{4}{216} = \frac{1}{54}$.
The probability that the sum of the numbers on the uppermost faces is at least $5$ is $P(\text{sum} \geq 5) = 1 - P(E) = 1 - \frac{1}{54} = \frac{53}{54}$.

(C) Total number of balls $= 6 + 4 = 10$.
Number of ways to draw $2$ balls from $10$ is given by $^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$.
For the balls to be of the same colour,they must either be both white or both black.
Number of ways to draw $2$ white balls from $6$ is $^{6}C_2 = \frac{6 \times 5}{2 \times 1} = 15$.
Number of ways to draw $2$ black balls from $4$ is $^{4}C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Total favorable outcomes $= 15 + 6 = 21$.
Required probability $= \frac{21}{45} = \frac{7}{15}$.

(A) Key Idea: Use the sine rule,i.e.,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Given,$\sin B \sin C = \frac{bc}{a^2}$.
From the sine rule,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation:
$\left(\frac{b}{2R}\right) \left(\frac{c}{2R}\right) = \frac{bc}{a^2}$
$\Rightarrow \frac{bc}{4R^2} = \frac{bc}{a^2}$
$\Rightarrow 4R^2 = a^2$
$\Rightarrow a = 2R$.
Since $\frac{a}{\sin A} = 2R$,we have $\frac{2R}{\sin A} = 2R$,which implies $\sin A = 1$.
Therefore,$A = 90^{\circ}$.
Thus,the triangle is a right-angled triangle.

(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$b = k \sin B$ and $c = k \sin C$.
Substituting these into the expression:
$\frac{b \sin B - c \sin C}{\sin (B - C)} = \frac{(k \sin B) \sin B - (k \sin C) \sin C}{\sin (B - C)}$
$= \frac{k(\sin^2 B - \sin^2 C)}{\sin (B - C)}$
Using the identity $\sin^2 B - \sin^2 C = \sin(B + C) \sin(B - C)$:
$= \frac{k \sin(B + C) \sin(B - C)}{\sin (B - C)}$
$= k \sin(B + C)$
Since $A + B + C = 180^{\circ}$,then $B + C = 180^{\circ} - A$.
$= k \sin(180^{\circ} - A) = k \sin A$
$= a$.

(D) Given $\cos A = \frac{\sin B}{\sin C}$.
Using the Sine Rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation: $\cos A = \frac{b/2R}{c/2R} = \frac{b}{c}$.
Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Equating the two expressions: $\frac{b^2 + c^2 - a^2}{2bc} = \frac{b}{c}$.
Multiplying both sides by $2bc$: $b^2 + c^2 - a^2 = 2b^2$.
Rearranging the terms: $c^2 = a^2 + b^2$.
Since the square of one side is equal to the sum of the squares of the other two sides,the triangle satisfies the Pythagorean theorem.
Therefore,$\triangle ABC$ is a right-angled triangle.

(C) We know that in $\triangle ABC$,$\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.
Given $\left(\tan \frac{A}{2}\right)\left(\tan \frac{B}{2}\right)=\frac{3}{4}$.
Substituting the formulas,we get $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \times \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \frac{3}{4}$.
Simplifying the expression,$\sqrt{\frac{(s-c)^2}{s^2}} = \frac{3}{4}$,which gives $\frac{s-c}{s} = \frac{3}{4}$.
Substituting $s = \frac{a+b+c}{2}$,we have $\frac{\frac{a+b+c}{2} - c}{\frac{a+b+c}{2}} = \frac{3}{4}$.
This simplifies to $\frac{a+b-c}{a+b+c} = \frac{3}{4}$.
Cross-multiplying gives $4(a+b) - 4c = 3(a+b) + 3c$.
Therefore,$a+b = 7c$.

(B) In any $\triangle ABC$,the area $\Delta$ is given by $\Delta = \frac{1}{2}bc \sin A$.
From this,we have $\sin A = \frac{2\Delta}{bc} \quad (i)$.
Also,by the sine rule,the circumradius $R$ is given by $R = \frac{a}{2 \sin A}$,which implies $\sin A = \frac{a}{2R} \quad (ii)$.
Equating $(i)$ and $(ii)$,we get $\frac{2\Delta}{bc} = \frac{a}{2R}$.
Solving for $\Delta$,we obtain $\Delta = \frac{abc}{4R}$.

(A) We are given the set $A = \{x \in \mathbb{R} : x^2 + 5|x| + 6 = 0\}$.
Since $x^2 = |x|^2$,the equation can be written as $|x|^2 + 5|x| + 6 = 0$.
Let $|x| = t$,where $t \ge 0$. The equation becomes $t^2 + 5t + 6 = 0$.
Factoring the quadratic,we get $(t + 2)(t + 3) = 0$.
This gives $t = -2$ or $t = -3$.
Since $t = |x|$ must be non-negative $(t \ge 0)$,there are no real values of $x$ that satisfy these conditions.
Therefore,the set $A$ is an empty set,i.e.,$A = \emptyset$.
Thus,the number of elements in $A$ is $n(A) = 0$.

(B) Let the first term be $a$ and the common ratio be $R$ of the $GP$.
Then,$A = aR^{p-1}$,$B = aR^{q-1}$,and $C = aR^{r-1}$.
Now,consider the expression $E = A^{q-r} \cdot B^{r-p} \cdot C^{p-q}$.
Substituting the values:
$E = (aR^{p-1})^{q-r} \cdot (aR^{q-1})^{r-p} \cdot (aR^{r-1})^{p-q}$
$E = a^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$
Since $(q-r) + (r-p) + (p-q) = 0$,the power of $a$ is $0$.
For the power of $R$:
$(pq - pr - q + r) + (qr - pq - r + p) + (rp - rq - p + q) = 0$
Thus,$E = a^0 \cdot R^0 = 1 \cdot 1 = 1$.

(C) Let the first term be $a$ and the common ratio be $r$.
Given that the sum of an infinite $GP$ is $S_{\infty} = \frac{a}{1-r} = 9$.
This implies $a = 9(1-r)$.
The sum of the first two terms is $a + ar = 5$.
Substituting $a = 9(1-r)$ into the equation:
$9(1-r)(1+r) = 5$.
$9(1-r^2) = 5$.
$1-r^2 = \frac{5}{9}$.
$r^2 = 1 - \frac{5}{9} = \frac{4}{9}$.
Thus,$r = \pm \frac{2}{3}$.
Since the options provided include $\frac{2}{3}$,the correct common ratio is $\frac{2}{3}$.

(B) In a $G.P.$,the $m^{\text{th}}$ term is the geometric mean of the terms that are equidistant from it.
Therefore,$(T_m)^2 = T_{m+n} \times T_{m-n}$.
Given $T_{m+n} = p$ and $T_{m-n} = q$.
Substituting these values,we get $(T_m)^2 = p \times q$.
Thus,$T_m = \sqrt{pq}$.

(C) We are given the sum of the first $n$ terms as $S_n = 5(2^n - 1)$.
We know that the $n^{th}$ term $t_n$ is given by $t_n = S_n - S_{n-1}$ for $n > 1$.
$t_n = 5(2^n - 1) - 5(2^{n-1} - 1)$
$t_n = 5(2^n - 1 - 2^{n-1} + 1)$
$t_n = 5(2^n - 2^{n-1})$
$t_n = 5(2 \times 2^{n-1} - 2^{n-1})$
$t_n = 5(2^{n-1}(2 - 1))$
$t_n = 5(2^{n-1})$

(A) Given the sum $\sum_{r=1}^n(2r+1) = 440$.
Expanding the summation,we get the series $3 + 5 + 7 + \ldots + (2n+1) = 440$.
This is an arithmetic progression with first term $a = 3$,common difference $d = 2$,and number of terms $n$.
The sum of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values: $\frac{n}{2}[2(3) + (n-1)(2)] = 440$.
$\Rightarrow \frac{n}{2}[6 + 2n - 2] = 440$.
$\Rightarrow \frac{n}{2}[2n + 4] = 440$.
$\Rightarrow n(n + 2) = 440$.
$\Rightarrow n^2 + 2n - 440 = 0$.
Solving the quadratic equation: $(n + 22)(n - 20) = 0$.
Since $n$ must be positive,$n = 20$.

(B) Given $S_n = \frac{4^n - 3^n}{3^n}$.
We know that $t_n = S_n - S_{n-1}$ for $n > 1$.
For $n = 2$,$t_2 = S_2 - S_1$.
Calculate $S_2 = \frac{4^2 - 3^2}{3^2} = \frac{16 - 9}{9} = \frac{7}{9}$.
Calculate $S_1 = \frac{4^1 - 3^1}{3^1} = \frac{4 - 3}{3} = \frac{1}{3}$.
Therefore,$t_2 = \frac{7}{9} - \frac{1}{3} = \frac{7 - 3}{9} = \frac{4}{9}$.

(B) Given set $A = \{x \mid x \in N, x \text{ is a prime number less than } 12\}$.
Since prime numbers less than $12$ are $2, 3, 5, 7, 11$,we have $A = \{2, 3, 5, 7, 11\}$.
Given set $B = \{x \mid x \in N, x \text{ is a factor of } 10\}$.
Since factors of $10$ are $1, 2, 5, 10$,we have $B = \{1, 2, 5, 10\}$.
The intersection $A \cap B$ consists of elements common to both sets $A$ and $B$.
$A \cap B = \{2, 3, 5, 7, 11\} \cap \{1, 2, 5, 10\} = \{2, 5\}$.

(D) Given the set $A = \{x \in R : x^2 - 5|x| + 6 = 0\}$.
Since $x^2 = |x|^2$,the equation becomes $|x|^2 - 5|x| + 6 = 0$.
Let $|x| = t$,then $t^2 - 5t + 6 = 0$.
Factoring the quadratic equation: $(t - 2)(t - 3) = 0$.
So,$|x| = 2$ or $|x| = 3$.
If $|x| = 2$,then $x = 2$ or $x = -2$.
If $|x| = 3$,then $x = 3$ or $x = -3$.
Thus,the set $A = \{-3, -2, 2, 3\}$.
The number of elements in set $A$ is $n(A) = 4$.

(D) The polar coordinates of $P$ are given as $\left(2, \frac{\pi}{6}\right)$.
If $Q$ is the reflection (image) of $P$ about the $X$-axis,the radial distance $r$ remains the same,and the angle $\theta$ changes to $-\theta$.
Therefore,the coordinates of $Q$ are $\left(2, -\frac{\pi}{6}\right)$.
To express the angle in the standard range $[0, 2\pi)$,we add $2\pi$ to the angle:
$-\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$.
Thus,the polar coordinates of $Q$ are $\left(2, \frac{11\pi}{6}\right)$.

(B) Given Cartesian coordinates $(x, y) = (-\sqrt{2}, \sqrt{2})$.
To find polar coordinates $(r, \theta)$:
$r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.
Since $x = r \cos \theta$ and $y = r \sin \theta$,we have:
$\cos \theta = \frac{-\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,the point lies in the second quadrant.
$\theta = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.
Thus,the polar coordinates are $\left(2, \frac{3 \pi}{4}\right)$.

(B) The median through vertex $R$ bisects the side $PQ$ at point $M$.
$M$ is the midpoint of $PQ$,so $M = \left( \frac{2 + (-2)}{2}, \frac{2 + 4}{2} \right) = (0, 3)$.
The median passes through $R(3, 4)$ and $M(0, 3)$.
The slope of the line $RM$ is $m = \frac{3 - 4}{0 - 3} = \frac{-1}{-3} = \frac{1}{3}$.
The equation of the line is $y - y_1 = m(x - x_1)$.
$y - 3 = \frac{1}{3}(x - 0)$
$3(y - 3) = x$
$3y - 9 = x$
$x - 3y + 9 = 0$.

(B) The given line is $x - 2y = 4$,which can be written as $2y = x - 4$ or $y = \frac{1}{2}x - 2$. The slope of this line is $m_1 = \frac{1}{2}$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$. Thus,$m_2 = -2$.
The equation of the line passing through $A(6, 1)$ with slope $m_2 = -2$ is given by $y - y_1 = m_2(x - x_1)$.
Substituting the values,we get $y - 1 = -2(x - 6)$.
$y - 1 = -2x + 12$.
$2x + y = 13$.
To find the $y$-intercept,we set $x = 0$ in the equation $2x + y = 13$.
$2(0) + y = 13 \Rightarrow y = 13$.
Therefore,the $y$-intercept of the line is $13$.

(D) The given lines are $x-3=0$ $(i)$ and $x+y=19$ (ii).
For line $(i)$,$x=3$,which is a vertical line parallel to the $y$-axis. The angle it makes with the positive $x$-axis is $\theta_1 = 90^{\circ}$.
For line (ii),$x+y=19$,we can rewrite it as $y = -x + 19$. Comparing this with the slope-intercept form $y = mx + c$,the slope $m_2 = -1$.
Since $m_2 = \tan \theta_2 = -1$,the angle $\theta_2 = 135^{\circ}$.
The angle between the two lines is $|\theta_2 - \theta_1| = |135^{\circ} - 90^{\circ}| = 45^{\circ}$.
Since $45^{\circ}$ is an acute angle,the required angle is $45^{\circ}$.

(D) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Given $A(7, -8, 1)$,$B(p, q, 5)$,$C(q+1, 5p, 0)$,and $G(3, -5, r)$.
Equating the coordinates:
$x$-coordinate: $\frac{7+p+q+1}{3} = 3 \Rightarrow p+q+8 = 9 \Rightarrow p+q = 1$ (Equation $1$)
$y$-coordinate: $\frac{-8+q+5p}{3} = -5 \Rightarrow 5p+q-8 = -15 \Rightarrow 5p+q = -7$ (Equation $2$)
$z$-coordinate: $\frac{1+5+0}{3} = r \Rightarrow r = \frac{6}{3} = 2$
Subtracting Equation $1$ from Equation $2$: $(5p+q) - (p+q) = -7 - 1 \Rightarrow 4p = -8 \Rightarrow p = -2$.
Substituting $p = -2$ into Equation $1$: $-2 + q = 1 \Rightarrow q = 3$.
Thus,the values are $p = -2, q = 3, r = 2$.

(D) The period of the function $\cos (kx)$ is given by $T = \frac{2 \pi}{|k|}$.
We check each option:
$A) \cos \left( \frac{\pi}{3} x \right) \implies T = \frac{2 \pi}{\pi/3} = 6$.
$B) \cos \left( \frac{\pi}{2} x \right) \implies T = \frac{2 \pi}{\pi/2} = 4$.
$C) \cos (2 \pi x) \implies T = \frac{2 \pi}{2 \pi} = 1$.
$D) \cos (\pi x) \implies T = \frac{2 \pi}{\pi} = 2$.
Therefore,the function with period $2$ is $\cos (\pi x)$.

(B) The range of the trigonometric functions are as follows:
$1$. For $\sin \theta$ and $\cos \theta$,the value must lie in the interval $[-1, 1]$.
$2$. For $\sec \theta$ and $\csc \theta$,the value must lie in $(-\infty, -1] \cup [1, \infty)$.
$3$. For $\tan \theta$ and $\cot \theta$,the value can be any real number $(R)$.
Evaluating the options:
- Option $A$: $\sec \theta = 23$ is possible since $23 \in (-\infty, -1] \cup [1, \infty)$.
- Option $B$: $\cos \theta = \sqrt{2} \approx 1.414$. Since $1.414 > 1$,this value lies outside the range $[-1, 1]$. Thus,$\cos \theta = \sqrt{2}$ has no solution.
- Option $C$: $\tan \theta = 2019$ is possible since the range of $\tan \theta$ is $R$.
- Option $D$: $\sin \theta = -\frac{1}{5} = -0.2$ is possible since $-0.2 \in [-1, 1]$.
Therefore,the correct option is $B$.

(C) Given the equation: $\sin^2 \theta = \frac{1}{2}$.
Taking the square root on both sides,we get $\sin \theta = \pm \frac{1}{\sqrt{2}}$.
Since $\theta \in [0, \pi]$,the sine function is non-negative in this interval.
Therefore,$\sin \theta = \frac{1}{\sqrt{2}}$.
The values of $\theta$ in $[0, \pi]$ that satisfy $\sin \theta = \frac{1}{\sqrt{2}}$ are $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$.
Thus,there are $2$ solutions.

(B) Given the equation: $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$: $2 \sin x \cos x = \frac{2}{4} = \frac{1}{2}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = \frac{1}{2}$.
Since $x \in \left(0, \frac{\pi}{2}\right)$,then $2x \in (0, \pi)$.
The solutions for $2x$ in $(0, \pi)$ are $2x = \frac{\pi}{6}$ and $2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Therefore,$x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$.

(B) Let $A = 18^{\circ}$. Then $5A = 90^{\circ}$,which implies $2A = 90^{\circ} - 3A$.
Taking sine on both sides,$\sin 2A = \sin(90^{\circ} - 3A) = \cos 3A$.
Using the double angle and triple angle formulas:
$2 \sin A \cos A = 4 \cos^3 A - 3 \cos A$.
Since $\cos 18^{\circ} \neq 0$,we can divide by $\cos A$:
$2 \sin A = 4 \cos^2 A - 3$.
Substituting $\cos^2 A = 1 - \sin^2 A$:
$2 \sin A = 4(1 - \sin^2 A) - 3$.
$2 \sin A = 4 - 4 \sin^2 A - 3$.
$4 \sin^2 A + 2 \sin A - 1 = 0$.
Using the quadratic formula for $\sin A$:
$\sin A = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.
Since $18^{\circ}$ is in the first quadrant,$\sin 18^{\circ} > 0$.
Therefore,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.

(D) Given $\theta = \frac{17 \pi}{3} = 5 \pi + \frac{2 \pi}{3}$.
Since $\tan(n \pi + x) = \tan x$ and $\cot(n \pi + x) = \cot x$,we have $\tan \theta = \tan \frac{2 \pi}{3} = -\sqrt{3}$ and $\cot \theta = \cot \frac{2 \pi}{3} = -\frac{1}{\sqrt{3}}$.
Now,calculate $(\tan \theta - \cot \theta) = -\sqrt{3} - (-\frac{1}{\sqrt{3}})$.
$= -\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{-3 + 1}{\sqrt{3}} = \frac{-2}{\sqrt{3}}$.

(A) In any $\triangle ABC$,the identity for the sum of tangents is $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Given that $\tan A + \tan B + \tan C = 6$,we have $\tan A \tan B \tan C = 6$.
We are also given $\tan A \cdot \tan B = 2$.
Substituting the value of $\tan A \cdot \tan B$ into the identity,we get $2 \cdot \tan C = 6$.
Therefore,$\tan C = \frac{6}{2} = 3$.

(C) Given the parametric equations of the curve are $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(t^{1/2}) = \frac{1}{2\sqrt{t}}$
$\frac{dy}{dt} = \frac{d}{dt}(t - t^{-1/2}) = 1 - (-\frac{1}{2})t^{-3/2} = 1 + \frac{1}{2t^{3/2}}$
Now,the slope of the tangent $\frac{dy}{dx}$ is given by $\frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{1 + \frac{1}{2t^{3/2}}}{\frac{1}{2\sqrt{t}}} = (1 + \frac{1}{2t^{3/2}}) \times (2\sqrt{t}) = 2\sqrt{t} + \frac{2\sqrt{t}}{2t^{3/2}} = 2\sqrt{t} + \frac{1}{t} = \frac{2t\sqrt{t} + 1}{t}$
At $t=4$,the slope of the tangent is:
$\left(\frac{dy}{dx}\right)_{t=4} = \frac{2(4)\sqrt{4} + 1}{4} = \frac{2(4)(2) + 1}{4} = \frac{16+1}{4} = \frac{17}{4}$
The slope of the normal is the negative reciprocal of the slope of the tangent:
$\text{Slope of normal} = -\frac{1}{(\frac{dy}{dx})_{t=4}} = -\frac{1}{17/4} = -\frac{4}{17}$

(B) Given the position function $x = 2 + 27t - t^3$.
To find the velocity,we differentiate $x$ with respect to $t$:
$v = \frac{dx}{dt} = 27 - 3t^2$.
The direction of motion reverses when the velocity becomes zero:
$27 - 3t^2 = 0$
$3t^2 = 27$
$t^2 = 9$
$t = 3$ (since $t > 0$).
Now,we calculate the distance $x$ at $t = 3$:
$x = 2 + 27(3) - (3)^3$
$x = 2 + 81 - 27$
$x = 56 \text{ units}$.

(A) Given the curve $y = \log_e x$.
First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at point $P(1, 0)$.
$\frac{dy}{dx} = \frac{1}{x}$.
At point $P(1, 0)$,the slope of the tangent $m_t = \left. \frac{dy}{dx} \right|_{(1, 0)} = \frac{1}{1} = 1$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$.
The equation of the normal line passing through $(x_1, y_1) = (1, 0)$ with slope $m_n = -1$ is:
$y - y_1 = m_n(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$
$x + y = 1$.

(A) Let the edge of the cube be $x \ cm$. The surface area $A$ of the cube is given by $A = 6x^2$.
Given that the rate of change of the edge is $\frac{dx}{dt} = -0.04 \ cm/sec$.
Differentiating $A$ with respect to $t$,we get $\frac{dA}{dt} = 12x \frac{dx}{dt}$.
Substituting the given values,$\frac{dA}{dt} = 12(10)(-0.04) = 120(-0.04) = -4.8 \ cm^2/sec$.
The negative sign indicates a decrease. Therefore,the rate of decrease of the surface area is $4.8 \ cm^2/sec$.

(A) Given function is $f(x) = x^2 - 2x + 1$.
We need to find the approximate value at $x = 2.99$.
Let $x = 3$ and $\Delta x = -0.01$,so that $x + \Delta x = 2.99$.
The derivative is $f'(x) = 2x - 2$.
Using the formula for linear approximation: $f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x)$.
At $x = 3$,$f(3) = 3^2 - 2(3) + 1 = 9 - 6 + 1 = 4$.
At $x = 3$,$f'(3) = 2(3) - 2 = 6 - 2 = 4$.
Substituting these values into the approximation formula:
$f(2.99) \approx f(3) + (-0.01) \cdot f'(3)$.
$f(2.99) \approx 4 + (-0.01)(4)$.
$f(2.99) \approx 4 - 0.04 = 3.96$.

(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \ cm/sec$.
At $t = 0$,the radius $r = 0$.
Integrating $\frac{dr}{dt} = 5$,we get $r = 5t$.
At $t = 2 \ seconds$,the radius $r = 5(2) = 10 \ cm$.
The area of the circular ripple is $A = \pi r^2$.
Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Substituting the values $r = 10 \ cm$ and $\frac{dr}{dt} = 5 \ cm/sec$:
$\frac{dA}{dt} = 2 \pi (10)(5) = 100 \pi \ cm^2/sec$.

(A) Given the function $f(x) = x^3 - 3x$.
First,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3$.
Factoring the derivative,we get:
$f'(x) = 3(x^2 - 1) = 3(x - 1)(x + 1)$.
To determine the intervals of increase and decrease,we find the critical points by setting $f'(x) = 0$:
$3(x - 1)(x + 1) = 0 \implies x = 1, x = -1$.
These points divide the real number line into three intervals: $(-\infty, -1)$,$(-1, 1)$,and $(1, \infty)$.
$1$. For $x \in (-\infty, -1)$,choose $x = -2$: $f'(-2) = 3((-2)^2 - 1) = 3(4 - 1) = 9 > 0$. Thus,$f(x)$ is increasing.
$2$. For $x \in (-1, 1)$,choose $x = 0$: $f'(0) = 3(0^2 - 1) = -3 < 0$. Thus,$f(x)$ is decreasing.
$3$. For $x \in (1, \infty)$,choose $x = 2$: $f'(2) = 3(2^2 - 1) = 3(3) = 9 > 0$. Thus,$f(x)$ is increasing.
Therefore,$f(x)$ is increasing in $(-\infty, -1) \cup (1, \infty)$ and decreasing in $(-1, 1)$.

(B) Given $f(x) = x + \frac{1}{x}$,$x \neq 0$.
Differentiating with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2}$.
For local maxima or minima,we set $f'(x) = 0$.
$1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1, -1$.
Now,we check the sign of $f'(x)$ around these points:
For $x < -1$,$f'(x) > 0$. For $-1 < x < 0$,$f'(x) < 0$. Since $f'(x)$ changes from positive to negative at $x = -1$,$x = -1$ is a point of local maxima.
The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -2$.
For $0 < x < 1$,$f'(x) < 0$. For $x > 1$,$f'(x) > 0$. Since $f'(x)$ changes from negative to positive at $x = 1$,$x = 1$ is a point of local minima.
The local minimum value is $f(1) = 1 + \frac{1}{1} = 2$.
Thus,the local maximum value is $-2$ and the local minimum value is $2$.

(B) Given the function $f(x) = 3x^3 - 9x^2 - 27x + 15$.
To find the critical points,we calculate the first derivative: $f'(x) = 9x^2 - 18x - 27$.
Setting $f'(x) = 0$,we get $9(x^2 - 2x - 3) = 0$,which factors to $9(x - 3)(x + 1) = 0$.
Thus,the critical points are $x = 3$ and $x = -1$.
To determine the nature of these points,we find the second derivative: $f''(x) = 18x - 18$.
At $x = 3$,$f''(3) = 18(3) - 18 = 36 > 0$,so $f(x)$ has a local minimum at $x = 3$.
At $x = -1$,$f''(-1) = 18(-1) - 18 = -36 < 0$,so $f(x)$ has a local maximum at $x = -1$.
The maximum value is $f(-1) = 3(-1)^3 - 9(-1)^2 - 27(-1) + 15 = -3 - 9 + 27 + 15 = 30$.

(C) The area $A$ is given by the integral of $|y|$ from $x=0$ to $x=\pi$.
Since $y = \cos x$ is positive in $[0, \pi/2]$ and negative in $[\pi/2, \pi]$,the area is:
$A = \int_0^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx \right|$
$A = [\sin x]_0^{\pi/2} + |[\sin x]_{\pi/2}^{\pi}|$
$A = (\sin(\pi/2) - \sin(0)) + |\sin(\pi) - \sin(\pi/2)|$
$A = (1 - 0) + |0 - 1|$
$A = 1 + |-1| = 1 + 1 = 2 \text{ sq. units.}$

(A) Given curves are $y=2x-x^2$ and $y=x$.
To find the points of intersection,set $2x-x^2 = x$.
$x-x^2 = 0 \implies x(1-x) = 0$.
Thus,the points of intersection are $x=0$ and $x=1$.
The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.
$\text{Area} = \int_0^1 (y_{\text{upper}} - y_{\text{lower}}) dx = \int_0^1 ((2x-x^2) - x) dx$.
$\text{Area} = \int_0^1 (x-x^2) dx$.
Evaluating the integral: $\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$.
$= (\frac{1}{2} - \frac{1}{3}) - (0 - 0) = \frac{3-2}{6} = \frac{1}{6}$ square units.

(A) We are given the objective function $Z = 5x + 4y$ subject to the constraints $y \leq 2x$,$x \leq 2y$,$x+y \leq 3$,$x \geq 0$,and $y \geq 0$.
First,we identify the feasible region by plotting the lines $y = 2x$,$x = 2y$,and $x+y = 3$.
$1$. The intersection of $y = 2x$ and $x+y = 3$: Substituting $y=2x$ into $x+y=3$ gives $x+2x=3$,so $3x=3$,which means $x=1$. Then $y=2(1)=2$. So,the point is $A(1, 2)$.
$2$. The intersection of $x = 2y$ and $x+y = 3$: Substituting $x=2y$ into $x+y=3$ gives $2y+y=3$,so $3y=3$,which means $y=1$. Then $x=2(1)=2$. So,the point is $B(2, 1)$.
$3$. The origin $O(0, 0)$ is also a corner point.
The feasible region is the triangle $OAB$. We evaluate $Z$ at the corner points:

Corner Point	$Z = 5x + 4y$
$O(0, 0)$	$5(0) + 4(0) = 0$
$A(1, 2)$	$5(1) + 4(2) = 5 + 8 = 13$
$B(2, 1)$	$5(2) + 4(1) = 10 + 4 = 14$

Thus,the maximum value of $Z$ is $14$ at point $B(2, 1)$.

(C) The objective function is $z = ax + by$ with $a, b > 0$. The constraints are $x \leq 2, y \leq 2, x + y \geq 3, x \geq 0, y \geq 0$.
Plotting these constraints,we identify the feasible region as the triangle with vertices $A(2, 1)$,$B(1, 2)$,and $C(2, 2)$.
Since the minimum value of $z$ occurs only at $(2, 1)$,the value of $z$ at $(2, 1)$ must be strictly less than the value of $z$ at the other corner points.
Comparing $z$ at $A(2, 1)$ and $B(1, 2)$:
$z(2, 1) = 2a + b$
$z(1, 2) = a + 2b$
For the minimum to be at $(2, 1)$,we must have $z(2, 1) < z(1, 2)$.
$2a + b < a + 2b$
$2a - a < 2b - b$
$a < b$
Thus,the correct condition is $a < b$.

(B) We are given the objective function: Maximize $z = 4x_1 + 2x_2$.
Subject to the constraints:
$1) 3x_1 + 2x_2 \geq 9$
$2) x_1 - x_2 \leq 3$
$3) x_1 \geq 0, x_2 \geq 0$
By plotting these lines on the coordinate plane:
For $3x_1 + 2x_2 = 9$,the intercepts are $(3, 0)$ and $(0, 4.5)$. The region is away from the origin.
For $x_1 - x_2 = 3$,the intercepts are $(3, 0)$ and $(0, -3)$. The region is towards the origin.
Upon analyzing the feasible region,we observe that the region is unbounded in the first quadrant. For an unbounded feasible region,if the objective function increases indefinitely as we move along the boundary,the $L$.$P$.$P$. has an unbounded solution. Here,as $x_1$ and $x_2$ increase,$z = 4x_1 + 2x_2$ can take arbitrarily large values within the feasible region. Therefore,the $L$.$P$.$P$. has an unbounded solution.

(A) To find the minimum value of the objective function $z = 10x + 25y$,we first identify the feasible region defined by the constraints $0 \leq x \leq 3$,$0 \leq y \leq 3$,and $x + y \geq 5$.
The vertices of the feasible region are determined by the intersection of the lines:
$1$. Intersection of $x = 3$ and $x + y = 5$ gives $y = 2$,so the point is $(3, 2)$.
$2$. Intersection of $x = 3$ and $y = 3$ gives the point $(3, 3)$.
$3$. Intersection of $y = 3$ and $x + y = 5$ gives $x = 2$,so the point is $(2, 3)$.
Now,we evaluate $z = 10x + 25y$ at these corner points:
- At $(3, 2)$: $z = 10(3) + 25(2) = 30 + 50 = 80$.
- At $(3, 3)$: $z = 10(3) + 25(3) = 30 + 75 = 105$.
- At $(2, 3)$: $z = 10(2) + 25(3) = 20 + 75 = 95$.
Comparing these values,the minimum value is $80$.

(D) The given constraints are $3x+2y \leq 12$,$2x+3y \leq 12$,$x \geq 0$,and $y \geq 0$.
To find the feasible region,we identify the corner points of the region bounded by these inequalities.
The lines are $L_1: 3x+2y=12$ and $L_2: 2x+3y=12$.
The intersection point $E$ of $L_1$ and $L_2$ is found by solving the system:
$3x+2y=12$ (multiply by $3$) $\Rightarrow 9x+6y=36$
$2x+3y=12$ (multiply by $2$) $\Rightarrow 4x+6y=24$
Subtracting gives $5x=12$,so $x=2.4$.
Substituting $x=2.4$ into $3(2.4)+2y=12$ gives $7.2+2y=12$,so $2y=4.8$,$y=2.4$.
The corner points of the feasible region are $(0,0)$,$(4,0)$,$(0,4)$,and $(2.4, 2.4)$.
Now evaluate $z=9x+11y$ at these points:
$z(0,0) = 9(0)+11(0) = 0$
$z(4,0) = 9(4)+11(0) = 36$
$z(0,4) = 9(0)+11(4) = 44$
$z(2.4, 2.4) = 9(2.4)+11(2.4) = 21.6+26.4 = 48$
The maximum value is $48$.

(C) Given that $f(x)$ is continuous at $x = 3$.
For a function to be continuous at a point $x = c$,the left-hand limit,right-hand limit,and the value of the function at that point must be equal.
$\therefore \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.
Calculating the left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) = \lim_{h \to 0} [a(3 - h) + 1] = 3a + 1$.
Calculating the right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) = \lim_{h \to 0} [b(3 + h) + 3] = 3b + 3$.
Since $f(x)$ is continuous at $x = 3$,we equate the limits: $3a + 1 = 3b + 3$.
Rearranging the terms: $3a - 3b = 3 - 1$.
$3(a - b) = 2$.
$a - b = \frac{2}{3}$.

(C) Given function is $f(x) = \begin{cases} x - \frac{|x|}{x}, & x < 0 \\ x + \frac{|x|}{x}, & x > 0 \\ 1, & x = 0 \end{cases}$.
To check continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(0)$.
$LHL = \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \left(x - \frac{|x|}{x}\right)$.
Since $x < 0$,$|x| = -x$,so $\frac{|x|}{x} = -1$.
$LHL = \lim_{x \to 0^{-}} (x - (-1)) = \lim_{x \to 0^{-}} (x + 1) = 0 + 1 = 1$.
$RHL = \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \left(x + \frac{|x|}{x}\right)$.
Since $x > 0$,$|x| = x$,so $\frac{|x|}{x} = 1$.
$RHL = \lim_{x \to 0^{+}} (x + 1) = 0 + 1 = 1$.
Given $f(0) = 1$.
Since $LHL = RHL = f(0) = 1$,the function $f(x)$ is continuous at $x = 0$.

(D) Given that the function $f(x)$ is continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Since $f(0) = 16$,we evaluate the limit:
$\lim_{x \to 0} \frac{(e^{kx} - 1) \tan kx}{4x^2} = 16$.
We can rewrite the expression as:
$\lim_{x \to 0} \left( \frac{e^{kx} - 1}{kx} \right) \left( \frac{\tan kx}{kx} \right) \left( \frac{k^2 x^2}{4x^2} \right) = 16$.
Using the standard limits $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\lim_{u \to 0} \frac{\tan u}{u} = 1$,we get:
$1 \times 1 \times \frac{k^2}{4} = 16$.
$\frac{k^2}{4} = 16$.
$k^2 = 64$.
$k = \pm 8$.

(B) Given that the function $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{\log(1 + ax) - \log(1 - bx)}{x}$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$f(0) = \lim_{x \to 0} \frac{\frac{d}{dx}(\log(1 + ax)) - \frac{d}{dx}(\log(1 - bx))}{\frac{d}{dx}(x)}$.
$f(0) = \lim_{x \to 0} \frac{\frac{a}{1 + ax} - \frac{-b}{1 - bx}}{1}$.
$f(0) = \lim_{x \to 0} \left( \frac{a}{1 + ax} + \frac{b}{1 - bx} \right)$.
Substituting $x = 0$:
$f(0) = \frac{a}{1 + 0} + \frac{b}{1 - 0} = a + b$.

(C) function $f(x)$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.
For option $C$,we check the limit at $x = 0$:
$R.H.L. = \lim_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \to 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} = \frac{1 - 0}{1 + 0} = 1$.
$L.H.L. = \lim_{x \to 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$.
Since $L.H.L. \neq R.H.L.$,the limit does not exist at $x = 0$.
Therefore,the function is not continuous at $x = 0$.

(D) Let $I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \cdot \sin^{3} \theta d \theta$
$= \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \cdot \sin \theta (1 - \cos^{2} \theta) d \theta$
Substitute $\cos \theta = t$,then $-\sin \theta d \theta = dt$,or $\sin \theta d \theta = -dt$.
When $\theta = 0, t = 1$ and when $\theta = \frac{\pi}{2}, t = 0$.
$I = \int_{1}^{0} \sqrt{t} (1 - t^{2}) (-dt) = \int_{0}^{1} (t^{1/2} - t^{5/2}) dt$
$= [\frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2}]_{0}^{1} = [\frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2}]_{0}^{1}$
$= (\frac{2}{3} - \frac{2}{7}) - 0 = \frac{14 - 6}{21} = \frac{8}{21}$

(C) Let $I = \int_0^4 \frac{1}{1+\sqrt{x}} \, dx$.
Substitute $x = t^2$,so $dx = 2t \, dt$.
When $x = 0, t = 0$ and when $x = 4, t = 2$.
Then,$I = \int_0^2 \frac{2t}{1+t} \, dt$.
$I = 2 \int_0^2 \frac{(1+t)-1}{1+t} \, dt$.
$I = 2 \int_0^2 \left(1 - \frac{1}{1+t}\right) \, dt$.
$I = 2 [t - \ln(1+t)]_0^2$.
$I = 2 [(2 - \ln 3) - (0 - \ln 1)]$.
$I = 2(2 - \ln 3) = 4 - 2\ln 3 = 4 - \ln(3^2) = 4 - \ln 9$.
Since $4 = \ln(e^4)$,we have $I = \ln(e^4) - \ln 9 = \ln \left(\frac{e^4}{9}\right)$.

(D) Let $I = \int_{0}^{a} \sqrt{\frac{a - x}{x}} dx$.
Substitute $x = a \sin^{2} \theta$,then $dx = 2a \sin \theta \cos \theta d \theta$.
When $x = 0$,$\theta = 0$. When $x = a$,$\theta = \frac{\pi}{2}$.
$I = \int_{0}^{\pi/2} \sqrt{\frac{a - a \sin^{2} \theta}{a \sin^{2} \theta}} (2a \sin \theta \cos \theta) d \theta$
$I = \int_{0}^{\pi/2} \frac{\cos \theta}{\sin \theta} (2a \sin \theta \cos \theta) d \theta$
$I = 2a \int_{0}^{\pi/2} \cos^{2} \theta d \theta$
Using the identity $\cos^{2} \theta = \frac{1 + \cos 2 \theta}{2}$,we get:
$I = 2a \int_{0}^{\pi/2} \frac{1 + \cos 2 \theta}{2} d \theta = a \int_{0}^{\pi/2} (1 + \cos 2 \theta) d \theta$
$I = a [\theta + \frac{\sin 2 \theta}{2}]_{0}^{\pi/2} = a [(\frac{\pi}{2} + 0) - (0 + 0)] = \frac{\pi a}{2}$.
Given $\int_{0}^{a} \sqrt{\frac{a - x}{x}} dx = \frac{K}{2}$,so $\frac{\pi a}{2} = \frac{K}{2}$.
Therefore,$K = \pi a$.

(B) Let $I = \int_{0}^{1} x(1 - x)^{5} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:
$I = \int_{0}^{1} (1 - x)(1 - (1 - x))^{5} dx$
$I = \int_{0}^{1} (1 - x)(x)^{5} dx$
$I = \int_{0}^{1} (x^{5} - x^{6}) dx$
Integrating term by term:
$I = \left[ \frac{x^{6}}{6} - \frac{x^{7}}{7} \right]_{0}^{1}$
$I = \left( \frac{1}{6} - \frac{1}{7} \right) - (0 - 0)$
$I = \frac{7 - 6}{42} = \frac{1}{42}$.

(B) Let $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} dx$ . . . .$(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we get:
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x)}} dx$
$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} dx$ . . . .$(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} dx$
$2I = \int_{a}^{b} 1 dx = [x]_{a}^{b} = b - a$
$I = \frac{b - a}{2}$

(B) Let $I = \int_{-3}^{3} (ax^5 + bx^3 + cx + k) dx$.
We can split the integral as:
$I = \int_{-3}^{3} (ax^5 + bx^3 + cx) dx + \int_{-3}^{3} k dx$.
Since $f(x) = ax^5 + bx^3 + cx$ is an odd function (i.e.,$f(-x) = -f(x)$),the integral of this function over the symmetric interval $[-3, 3]$ is $0$.
Therefore,$I = 0 + \int_{-3}^{3} k dx = \int_{-3}^{3} k dx$.
Evaluating this,we get $I = [kx]_{-3}^{3} = k(3) - k(-3) = 3k + 3k = 6k$.
Thus,the value of the integral depends only on the constant $k$.

(D) Given that $A$ and $B$ are square matrices of order $n=3$.
We are given $|A|=2$ and $|B|=4$.
We need to find $|A(\operatorname{adj} B)|$.
Using the property of determinants,$|AB| = |A||B|$,we have:
$|A(\operatorname{adj} B)| = |A| |\operatorname{adj} B|$.
We know that for a square matrix $B$ of order $n$,$|\operatorname{adj} B| = |B|^{n-1}$.
Here $n=3$,so $|\operatorname{adj} B| = |B|^{3-1} = |B|^2$.
Substituting the values:
$|A(\operatorname{adj} B)| = |A| \times |B|^2 = 2 \times (4)^2$.
$|A(\operatorname{adj} B)| = 2 \times 16 = 32$.

(C) The equation of a circle that lies in the first quadrant and touches both the axes is given by $(x - a)^2 + (y - a)^2 = a^2$,where $a$ is the radius of the circle.
This simplifies to $x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$,which is $x^2 + y^2 - 2ax - 2ay + a^2 = 0$.
Here,there is only one arbitrary constant,$a$.
The order of the differential equation is equal to the number of independent arbitrary constants in the general equation of the family of curves.
Since there is only $1$ arbitrary constant,the order of the differential equation is $1$.

(B) The general equation of a circle with radius $r=4$ and center $(h, k)$ is given by $(x-h)^2 + (y-k)^2 = 4^2$.
Here,$h$ and $k$ are arbitrary constants.
Since there are $2$ independent arbitrary constants,the order of the differential equation is equal to the number of arbitrary constants.
Therefore,the order of the differential equation is $2$.

(B) The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.
Given the center is $A(-1, 2)$,the equation becomes $(x - (-1))^2 + (y - 2)^2 = r^2$.
Expanding this,we get $(x + 1)^2 + (y - 2)^2 = r^2$.
$x^2 + 2x + 1 + y^2 - 4y + 4 = r^2$.
$x^2 + y^2 + 2x - 4y + 5 - r^2 = 0$.
Let $c = 5 - r^2$,where $c$ is an arbitrary constant.
Thus,the general solution is $x^2 + y^2 + 2x - 4y + c = 0$.

(D) Given differential equation is $(x^2 + 1) \frac{dy}{dx} + (y^2 + 1) = 0$.
Rearranging the terms to separate the variables,we get:
$(x^2 + 1) \frac{dy}{dx} = -(y^2 + 1)$
$\frac{dy}{y^2 + 1} = -\frac{dx}{x^2 + 1}$
Integrating both sides:
$\int \frac{dy}{y^2 + 1} = -\int \frac{dx}{x^2 + 1}$
We know that $\int \frac{du}{u^2 + 1} = \tan^{-1} u + C$.
Therefore,$\tan^{-1} y = -\tan^{-1} x + C$
$\tan^{-1} x + \tan^{-1} y = C$.

(A) Given the differential equation: $\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k$ is a constant.
Separating the variables: $\frac{d\theta}{\theta - \theta_0} = -k dt$.
Integrating both sides: $\int \frac{d\theta}{\theta - \theta_0} = \int -k dt$.
This gives: $\ln|\theta - \theta_0| = -kt + C_1$.
Taking the exponential of both sides: $\theta - \theta_0 = e^{-kt + C_1} = e^{C_1} e^{-kt}$.
Let $e^{C_1} = a$.
Then,$\theta - \theta_0 = a e^{-kt}$.
Therefore,the solution is $\theta = \theta_0 + a e^{-kt}$.

(D) Given differential equation is $\log\left(\frac{dy}{dx}\right) = x$.
By definition of logarithm,we have $\frac{dy}{dx} = e^x$.
Separating the variables,we get $dy = e^x dx$.
Integrating both sides,we get $\int dy = \int e^x dx$,which gives $y = e^x + C$.
Given that when $x = 0, y = 1$,substituting these values into the equation: $1 = e^0 + C$.
Since $e^0 = 1$,we have $1 = 1 + C$,which implies $C = 0$.
Therefore,the particular solution is $y = e^x$.

(C) The given differential equation is $x \frac{dy}{dx} = y - x \tan \left( \frac{y}{x} \right)$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} - \tan \left( \frac{y}{x} \right)$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v - \tan v$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = - \tan v$.
Separating the variables: $\frac{dv}{\tan v} = - \frac{dx}{x}$,which is $\cot v \, dv = - \frac{dx}{x}$.
Integrating both sides: $\int \cot v \, dv = - \int \frac{1}{x} \, dx$.
This gives $\ln |\sin v| = - \ln |x| + \ln |c|$.
Using logarithmic properties: $\ln |\sin v| + \ln |x| = \ln |c| \Rightarrow \ln |x \sin v| = \ln |c|$.
Thus,$x \sin v = c$. Substituting $v = \frac{y}{x}$ back,we get $x \sin \left( \frac{y}{x} \right) = c$.

(B) Given differential equation is $y \ dx - x \ dy = xy \ dx$.
Dividing both sides by $xy$,we get:
$\frac{y \ dx - x \ dy}{xy} = dx$
This can be written as:
$d \left( \log \left( \frac{x}{y} \right) \right) = dx$
Integrating both sides:
$\int d \left( \log \left( \frac{x}{y} \right) \right) = \int dx$
$\log \left( \frac{x}{y} \right) = x + C$
Assuming the constant of integration $C = 0$ for the given options:
$\log \left( \frac{x}{y} \right) = x$
$\frac{x}{y} = e^x$
$x = y e^x$

(B) Given $x = \sin \theta$ and $y = \sin^3 \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = \cos \theta$
$\frac{dy}{d\theta} = 3 \sin^2 \theta \cos \theta$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \sin^2 \theta \cos \theta}{\cos \theta} = 3 \sin^2 \theta$
Next,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3 \sin^2 \theta) = \frac{d}{d\theta}(3 \sin^2 \theta) \cdot \frac{d\theta}{dx}$
$\frac{d^2 y}{dx^2} = (6 \sin \theta \cos \theta) \cdot \frac{1}{\cos \theta} = 6 \sin \theta$
Finally,evaluate at $\theta = \frac{\pi}{2}$:
$\frac{d^2 y}{dx^2} = 6 \sin(\frac{\pi}{2}) = 6(1) = 6$.

(C) Given equation is $x^y = e^{x - y}$.
Taking $\log$ on both sides,we get $y \log x = (x - y) \log e = x - y$ . . . . . . $(i)$.
When $x = 1$,substituting in $(i)$ gives $y \log 1 = 1 - y$,which implies $0 = 1 - y$,so $y = 1$.
Differentiating both sides of $(i)$ with respect to $x$ using the product rule:
$y \cdot (\frac{1}{x}) + \log x \cdot \frac{dy}{dx} = 1 - \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} (\log x + 1) = 1 - \frac{y}{x}$.
$\frac{dy}{dx} = \frac{x - y}{x (\log x + 1)}$.
At $x = 1$ and $y = 1$:
$\frac{dy}{dx} = \frac{1 - 1}{1 (\log 1 + 1)} = \frac{0}{1} = 0$.

(A) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.
Squaring both sides,we get $x^2 = a^{\sin^{-1} t}$ and $y^2 = a^{\cos^{-1} t}$.
Taking the logarithm on both sides,we have $\log_a(x^2) = \sin^{-1} t$ and $\log_a(y^2) = \cos^{-1} t$.
Adding these two equations,we get $\log_a(x^2) + \log_a(y^2) = \sin^{-1} t + \cos^{-1} t$.
Using the property $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have $\log_a(x^2 y^2) = \frac{\pi}{2}$.
This implies $x^2 y^2 = a^{\pi/2}$,which is a constant.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(a^{\pi/2})$.
Using the product rule,$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$.
Dividing by $2xy$,we get $y + x \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = \frac{-y}{x}$.

(B) Given $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$.
Rationalizing the denominator inside the logarithm:
$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})(x + \sqrt{x^2 + 25})}{(\sqrt{x^2 + 25} - x)(\sqrt{x^2 + 25} + x)} \right]$
$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{(x^2 + 25) - x^2} \right]$
$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{25} \right]$
Using logarithmic properties:
$y = 2 \log (x + \sqrt{x^2 + 25}) - \log 25$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{x + \sqrt{x^2 + 25}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + 25}) - 0$
$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( 1 + \frac{1}{2\sqrt{x^2 + 25}} \cdot 2x \right)$
$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( \frac{\sqrt{x^2 + 25} + x}{\sqrt{x^2 + 25}} \right)$
$\frac{dy}{dx} = \frac{2}{\sqrt{x^2 + 25}}$

(B) Let $y = \log _{e^2}(\log x)$.
Using the logarithmic property $\log _{b^n} a = \frac{1}{n} \log _b a$,we can rewrite the expression as:
$y = \frac{1}{2} \log _e(\log x)$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left[ \frac{1}{2} \log _e(\log x) \right]$
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)$
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\log x} \cdot \frac{1}{x}$
$\frac{dy}{dx} = \frac{1}{2x \log x}$
Thus,the correct option is $B$.

(A) Let $y = \sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$.
Put $t = \tan \theta$,then $\theta = \tan ^{-1} t$.
$y = \sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right) = \sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) = \sin ^{-1}(\sin \theta) = \theta = \tan ^{-1} t$.
Let $z = \cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$.
$z = \cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right) = \cos ^{-1}\left(\frac{1}{\sec \theta}\right) = \cos ^{-1}(\cos \theta) = \theta = \tan ^{-1} t$.
Since $y = \theta$ and $z = \theta$,we have $y = z$.
Therefore,$\frac{dy}{dz} = \frac{d}{dz}(z) = 1$.

(A) Given $f(x) = \cos^{-1} \left( \frac{1 - (\log x)^2}{1 + (\log x)^2} \right)$.
Let $u = \log x$. Then the expression becomes $f(x) = \cos^{-1} \left( \frac{1 - u^2}{1 + u^2} \right)$.
Using the trigonometric substitution $u = \tan \theta$,we know that $\cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) = \cos^{-1} (\cos 2\theta) = 2\theta = 2 \tan^{-1} u$.
Thus,$f(x) = 2 \tan^{-1} (\log x)$.
Differentiating with respect to $x$ using the chain rule:
$f'(x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{d}{dx}(\log x) = \frac{2}{1 + (\log x)^2} \cdot \frac{1}{x}$.
Now,evaluating at $x = e$:
$f'(e) = \frac{2}{1 + (\log e)^2} \cdot \frac{1}{e} = \frac{2}{1 + 1^2} \cdot \frac{1}{e} = \frac{2}{2} \cdot \frac{1}{e} = \frac{1}{e}$.

(B) For the function $f(x) = \sqrt{\frac{x-2}{3-x}}$ to be defined,the expression under the square root must be non-negative: $\frac{x-2}{3-x} \geq 0$.
This implies that the numerator and denominator must have opposite signs or the numerator must be zero.
Specifically,$x-2 \geq 0$ and $3-x > 0$.
From $x-2 \geq 0$,we get $x \geq 2$.
From $3-x > 0$,we get $x < 3$.
Combining these two conditions,we get $2 \leq x < 3$.
Thus,the domain of $f(x)$ is $[2, 3)$.

(C) We have $f(x) = [x]$.
By the definition of the right-hand derivative at $x = 1$:
$f^{\prime}(1^{+}) = \lim_{h \to 0^{+}} \frac{f(1+h) - f(1)}{h}$.
Since $h$ is a small positive value,$1+h$ is slightly greater than $1$,so $[1+h] = 1$.
Also,$[1] = 1$.
Substituting these values:
$f^{\prime}(1^{+}) = \lim_{h \to 0^{+}} \frac{1 - 1}{h} = \lim_{h \to 0^{+}} \frac{0}{h} = 0$.

(D) Given functions are $f(x)=3x+6$ and $g(x)=4x+k$.
Since $f \circ g(x) = g \circ f(x)$,we have:
$f(g(x)) = g(f(x))$
$f(4x+k) = g(3x+6)$
Substitute the expressions into the functions:
$3(4x+k)+6 = 4(3x+6)+k$
$12x + 3k + 6 = 12x + 24 + k$
Subtract $12x$ from both sides:
$3k + 6 = 24 + k$
Rearrange the terms to solve for $k$:
$3k - k = 24 - 6$
$2k = 18$
$k = 9$

(A) Given: $f(x) = 3x - 2$ and $g(x) = x^2$.
By the definition of composite function,$f \circ g(x) = f(g(x))$.
Substitute $g(x) = x^2$ into the function $f(x)$:
$f(g(x)) = f(x^2)$.
Since $f(x) = 3x - 2$,replace $x$ with $x^2$:
$f(x^2) = 3(x^2) - 2 = 3x^2 - 2$.
Therefore,$f \circ g(x) = 3x^2 - 2$.

(D) The surface area $S$ of a spherical balloon with radius $r$ is given by $S = 4 \pi r^2$.
According to the problem,the rate of change of the surface area is constant,$K$.
Therefore,$\frac{d}{dt}(4 \pi r^2) = K$.
Integrating both sides with respect to $t$:
$\int \frac{d}{dt}(4 \pi r^2) dt = \int K dt$.
This gives $4 \pi r^2 = K t + C$,where $C$ is the constant of integration.

(D) Let $I = \int \frac{dx}{(\sin x + \cos x)(2 \cos x + \sin x)}$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x}{(\tan x + 1)(2 + \tan x)} dx$.
Let $\tan x = t$,then $\sec^2 x dx = dt$.
$I = \int \frac{dt}{(t+1)(t+2)}$.
Using partial fractions: $\frac{1}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$.
$1 = A(t+2) + B(t+1)$.
For $t = -1$,$A = 1$. For $t = -2$,$B = -1$.
$I = \int \left( \frac{1}{t+1} - \frac{1}{t+2} \right) dt = \log|t+1| - \log|t+2| + C$.
$I = \log \left| \frac{t+1}{t+2} \right| + C = \log \left| \frac{\tan x + 1}{\tan x + 2} \right| + C$.

(A) Let $I = \int \frac{1}{1-\cot x} dx$.
Substitute $\cot x = \frac{\cos x}{\sin x}$,so $I = \int \frac{1}{1-\frac{\cos x}{\sin x}} dx = \int \frac{\sin x}{\sin x - \cos x} dx$.
Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} dx$.
Rewrite the numerator: $2 \sin x = (\sin x - \cos x) + (\sin x + \cos x)$.
Thus,$I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} dx = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) dx$.
Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) dx$.
$I = \frac{1}{2} \left( \int 1 dx + \int \frac{1}{u} du \right) = \frac{1}{2} (x + \log |u|) + C = \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C$.
Comparing with $Ax + B \log |\sin x - \cos x| + C$,we get $A = \frac{1}{2}$ and $B = \frac{1}{2}$.
Therefore,$A + B = \frac{1}{2} + \frac{1}{2} = 1$.